A pendulum clock keeps correct time at `20^(@)C` and coefficient of linear expansion of pendulum is `12xx10^(-6)//.^(@)C`. If the room temperature increases to `40^(@)C`, how many seconds the clock lose or gain per day?

Options

A. 10.36 s
B. 20.6 s
C. 5 s
D. 20 minutes

Answer

A

Explanation

`T=2pisqrt((l)/(g))T^(2)=(4pi^(2)l)/(g)`
`T^(2)=Kl" ... (i)"`
Where `K=(4pi^(2))/(g)`
Differentiating equation (i),
2TdT = Kdl `" … (ii)"`
From equation (i) and (ii),
`therefore (dT)/(T)=(kdl)/(2l)" ... (iii)"`
But `alpha=(dl)/(LDeltat) therefore alphaDeltat=(dl)/(L)`
Hence equation (iii) becomes,
`(dT)/(T)=(1)/(2)alphaDeltat`
`=(1)/(2)xx12xx10^(-6)xx(40-20)`
`=12xx10^(-5)`
The period of clock pendulum T = 1 s
`therefore dT=12xx10^(-5)s`
Thus the clock gains `12xx10^(-5)` s every osci.
`therefore` In one day (86400 osci.) the clock will gain time of `12xx10^(-5)xx86400=10.37s`